Reciprocal Function is Unbounded on Open Unit Interval

Theorem

Let $A = \openint 0 1$ denote the open unit interval.

Let $f: A \to \R$ be the reciprocal function:

$\forall x \in A: \map f x := \dfrac 1 x$

Then $f$ is unbounded.

Proof

Let $K \in \R_{>0}$.

Then:

$\exists x \in \R: 0 < x < \dfrac 1 K$ such that $x < 1$.

Then we have:

$\map f x = \dfrac 1 x > K$

So whatever $K$ may be, it can never be large enough to be a bound of $f$ on $\openint 0 1$.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.1$: Motivation: Example $5.1.2$